In the following,the correct bond order seque | vedclass

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Question

(A) According to Molecular Orbital Theory $(MOT)$,the bond order is calculated as: $	ext{Bond Order} = \frac{N_b - N_a}{2}$,where $N_b$ is the number

Vedclass · Accepted Answer

$O_{2}^{+} > O_{2} > O_{2}^{-} > O_{2}^{2-}$

Vedclass · Answer

(A) According to Molecular Orbital Theory $(MOT)$,the bond order is calculated as: $	ext{Bond Order} = \frac{N_b - N_a}{2}$,where $N_b$ is the number of bonding electrons and $N_a$ is the number of antibonding electrons.For $O_{2}^{+} (15 \ e^-)$: $	ext{Bond Order} = \frac{10 - 5}{2} = 2.5$For $O_{2} (16 \ e^-)$: $	ext{Bond Order} = \frac{10 - 6}{2} = 2.0$For $O_{2}^{-} (17 \ e^-)$: $	ext{Bond Order} = \frac{10 - 7}{2} = 1.5$For $O_{2}^{2-} (18 \ e^-)$: $	ext{Bond Order} = \frac{10 - 8}{2} = 1.0$Thus,the correct sequence is $O_{2}^{+} > O_{2} > O_{2}^{-} > O_{2}^{2-}$.

In the following,the correct bond order sequence is:

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